By invariance of domain, the condition $\dim V=\dim_pX$ implies that the local injectivity is locally a bijection to a neighborhood of $p.$ So there are open sets $V_0\subseteq V$ and $X_0\subseteq X$ with $V_0$ containing $0,$ and a function $g:V_0\to X_0$ such that $\pi_V(g(v)-p)=v$ and $g(0)=p.$ By condition 2 we have $g(v)=p+v+o(\|v\|)$ with implicit constants depending only on $X$ and $p.$

Assume there's another $V'$ satisfying the same conditions. For each $v\in V$ we have $\lim_{\lambda\to 0^+}\frac{\|\pi_{V'^\perp}(g(\lambda v)-p)\|}{\lambda \|v\|}=0.$ Using $g(\lambda v)=p+\lambda v+o(\lambda)\|v\|$ we get $\pi_{V'^\perp}(v)=0,$ so $v\in V'.$ Since $\dim V=\dim V'$ we must have $V=V'.$

By invariance of domain, the condition $\dim V=\dim_pX$ implies that the local injectivity is locally a bijection to a neighborhood of $p.$ So there are open sets $V_0\subseteq V$ and $X_0\subseteq X$ with $V_0$ containing $0,$ and a function $g:V_0\to X_0$ such that $\pi_V(g(v)-p)=v$ and $g(0)=p.$ By condition 2 we have $g(v)=p+v+o(\|v\|)$ with implicit constants depending only on $X,V$ and $p.$

Assume there's another $V'$ satisfying the same conditions. For each $v\in V$ we have $\lim_{\lambda\to 0^+}\frac{\|\pi_{V'^\perp}(g(\lambda v)-p)\|}{\|g(\lambda v)-p\|}=0.$ Using $g(\lambda v)=p+\lambda v+o(\lambda\|v\|)$ we get $\pi_{V'^\perp}(v)=0,$ so $v\in V'.$ Since $\dim V=\dim V'$ we must have $V=V'.$